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Question Number 31076 by abdo imad last updated on 02/Mar/18

find  ∫_0 ^((√2)/2)        (dx/((2x^2  +1)(√(1+x^2 )))) .

$${find}\:\:\int_{\mathrm{0}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\:\:\:\:\:\:\frac{{dx}}{\left(\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}\right)\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:. \\ $$

Commented by abdo imad last updated on 07/Mar/18

the ch.x=sht give   I= ∫_0 ^(argsh(((√2)/2)))     ((cht dt)/((2 sh^2 t +1)cht))=∫_0 ^(argsh(((√2)/2)))     (dt/(2sh^2 t +1))  sh^2 t= (((e^t  −e^(−t) )/2))^2 =(1/4)(e^(2t)  +e^(−2t)  −2)⇒2sh^2 t=(1/2)( e^(2t)  +e^(−2t) −2)  =((e^(2t)  +e^(−2t) )/2) −1 ⇒ 2sh^2 t +1 = ((e^(2t)  +e^(−2t) )/2) ⇒  I = ∫_0 ^(argsh(((√2)/2)))   ((2dt)/(e^(2t)  +e^(−2t) ))  let use tbe chan. e^(2t) =u ⇒2t=lnu  I= ∫_1 ^e^(2argh(((√2)/2)))    (2/(u +(1/u)))  (du/(2u))=∫_1 ^e^(2argsh(((√2)/2)))   (du/(1+u^2 ))=[arctanu]_1 ^e^(2argsh(((√2)/2)))     but  we have argshx=ln(x+(√(1+x^2 ))) ⇒  argsh(((√2)/2))=ln( ((√2)/2) +(√(3/2)) )⇒2argsh(((√2)/2))=ln((((√2)/2) +((√3)/(√2)))^2 )  and e^(2argsh(((√2)/2))) = (1/2) +(√3) +(3/2)=2+(√3)  ⇒  I= arctan(2+(√3)) −(π/4) .

$${the}\:{ch}.{x}={sht}\:{give}\: \\ $$$${I}=\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} \:\:\:\:\frac{{cht}\:{dt}}{\left(\mathrm{2}\:{sh}^{\mathrm{2}} {t}\:+\mathrm{1}\right){cht}}=\int_{\mathrm{0}} ^{{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} \:\:\:\:\frac{{dt}}{\mathrm{2}{sh}^{\mathrm{2}} {t}\:+\mathrm{1}} \\ $$$${sh}^{\mathrm{2}} {t}=\:\left(\frac{{e}^{{t}} \:−{e}^{−{t}} }{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left({e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:−\mathrm{2}\right)\Rightarrow\mathrm{2}{sh}^{\mathrm{2}} {t}=\frac{\mathrm{1}}{\mathrm{2}}\left(\:{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} −\mathrm{2}\right) \\ $$$$=\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:−\mathrm{1}\:\Rightarrow\:\mathrm{2}{sh}^{\mathrm{2}} {t}\:+\mathrm{1}\:=\:\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} \:\:\frac{\mathrm{2}{dt}}{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }\:\:{let}\:{use}\:{tbe}\:{chan}.\:{e}^{\mathrm{2}{t}} ={u}\:\Rightarrow\mathrm{2}{t}={lnu} \\ $$$${I}=\:\int_{\mathrm{1}} ^{{e}^{\mathrm{2}{argh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} } \:\:\frac{\mathrm{2}}{{u}\:+\frac{\mathrm{1}}{{u}}}\:\:\frac{{du}}{\mathrm{2}{u}}=\int_{\mathrm{1}} ^{{e}^{\mathrm{2}{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} } \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\left[{arctanu}\right]_{\mathrm{1}} ^{{e}^{\mathrm{2}{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} } \:\:\:{but} \\ $$$${we}\:{have}\:{argshx}={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)={ln}\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\sqrt{\frac{\mathrm{3}}{\mathrm{2}}}\:\right)\Rightarrow\mathrm{2}{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)={ln}\left(\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{3}}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} \right) \\ $$$${and}\:{e}^{\mathrm{2}{argsh}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)} =\:\frac{\mathrm{1}}{\mathrm{2}}\:+\sqrt{\mathrm{3}}\:+\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{2}+\sqrt{\mathrm{3}}\:\:\Rightarrow \\ $$$${I}=\:{arctan}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:−\frac{\pi}{\mathrm{4}}\:. \\ $$

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